Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q, x, y, z, par[20][100100], b[100], h[100100], ta[30];
vector<int> v[100100], p[20][100100], a;
void koch(int xx, int yy, int zz) {
  int xxx = 0, yyy = 0;
  for (int i = 0; i < 10; i++) {
    if (xxx == p[zz][xx].size() && yyy == p[zz][yy].size()) break;
    if (xxx == p[zz][xx].size()) {
      p[zz + 1][xx].push_back(p[zz][yy][yyy++]);
      continue;
    }
    if (yyy == p[zz][yy].size()) {
      p[zz + 1][xx].push_back(p[zz][xx][xxx++]);
      continue;
    }
    if (p[zz][xx][xxx] >= p[zz][yy][yyy])
      p[zz + 1][xx].push_back(p[zz][yy][yyy++]);
    else
      p[zz + 1][xx].push_back(p[zz][xx][xxx++]);
  }
}
void dfs(int xx, int parpar, int k) {
  par[0][xx] = parpar;
  h[xx] = k;
  for (int i = 1; i < 20; i++) {
    par[i][xx] = par[i - 1][par[i - 1][xx]];
  }
  for (int i = 1; i < 20; i++) {
    koch(xx, par[i - 1][xx], i - 1);
  }
  for (int i = 0; i < v[xx].size(); i++) {
    if (parpar != v[xx][i]) {
      dfs(v[xx][i], xx, k + 1);
    }
  }
}
void koch1(int xx, int zz) {
  int xxx = 0, yyy = 0;
  for (int i = 0; i < 10; i++) {
    if (xxx == a.size() && yyy == p[zz][xx].size()) break;
    if (xxx == a.size()) {
      b[i] = p[zz][xx][yyy++];
      continue;
    }
    if (yyy == p[zz][xx].size()) {
      b[i] = (a[xxx++]);
      continue;
    }
    if (a[xxx] >= p[zz][xx][yyy])
      b[i] = (p[zz][xx][yyy++]);
    else
      b[i] = (a[xxx++]);
  }
  a.clear();
  for (int i = 0; i < 10; i++) {
    if (b[i] == 0) break;
    a.push_back(b[i]);
    b[i] = 0;
  }
}
int ipar(int xx, int hh) {
  for (int i = 19; i >= 0; i--) {
    if (hh >= ta[i]) {
      hh -= ta[i];
      koch1(xx, i);
      xx = par[i][xx];
    }
  }
  return xx;
}
void lca(int xx, int yy) {
  for (int i = 19; i >= 0; i--) {
    if (par[i][xx] != par[i][yy]) {
      koch1(xx, i);
      koch1(yy, i);
      xx = par[i][xx];
      yy = par[i][yy];
    }
  }
  if (xx == yy) {
    koch1(xx, 0);
    return;
  } else {
    koch1(xx, 1);
    koch1(yy, 0);
    return;
  }
}
void solve(int xx, int yy, int tt) {
  fill(b, b + 20, 0);
  a.clear();
  if (h[xx] > h[yy]) swap(xx, yy);
  yy = ipar(yy, h[yy] - h[xx]);
  lca(xx, yy);
  z = a.size();
  cout << min(tt, z) << ' ';
  for (int i = 0; i < min(tt, z); i++) {
    cout << a[i] << ' ';
  }
  cout << endl;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ta[0] = 1;
  for (int i = 1; i <= 21; i++) {
    ta[i] = ta[i - 1] * 2;
  }
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = 0; i < m; i++) {
    cin >> x;
    p[0][x].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) {
    sort(p[0][i + 1].begin(), p[0][i + 1].end());
  }
  dfs(1, 0, 1);
  for (int i = 0; i < q; i++) {
    cin >> x >> y >> m;
    solve(x, y, m);
  }
  return 0;
}